case study sequence and series class 11

CBSE Case Study Questions for Class 11 Maths Sequences and Series Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 11 Maths Sequences and Series  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 11 Maths Sequences and Series PDF

Checkout our case study questions for other chapters.

• Chapter 7 Permutations and Combinations Case Study Questions
• Chapter 8 Binomial Theorem Case Study Questions
• Chapter 10 Straight Lines Case Study Questions
• Chapter 11 Conic Sections Case Study Questions

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Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

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Class 11 Mathematics Case Study Questions

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The rationale behind teaching Mathematics

The general rationale to teach Mathematics at the senior secondary level is to assist students:

• In knowledge acquisition and cognitive understanding of basic ideas, words, principles, symbols, and mastery of underlying processes and abilities, notably through motivation and visualization.
• To experience the flow of arguments while demonstrating a point or addressing an issue.
• To use the information and skills gained to address issues using several methods wherever possible.
• To cultivate a good mentality in order to think, evaluate, and explain coherently.
• To spark interest in the subject by taking part in relevant tournaments.
• To familiarise pupils with many areas of mathematics utilized in daily life.
• To pique students’ interest in studying mathematics as a discipline.

Case studies in Class 11 Mathematics

A case study in mathematics is a comprehensive examination of a specific mathematical topic or scenario. Case studies are frequently used to investigate the link between theory and practise, as well as the connections between different fields of mathematics. A case study will frequently focus on a specific topic or circumstance and will investigate it using a range of methodologies. These approaches may incorporate algebraic, geometric, and/or statistical analysis.

Sample Class 11 Mathematics case study questions

When it comes to preparing for Class 11 Mathematics, one of the best things Class 11 Mathematics students can do is to look at some Class 11 Mathematics sample case study questions. Class 11 Mathematics sample case study questions will give you a good idea of the types of Class 11 Mathematics sample case study questions that will be asked in the exam and help you to prepare more effectively.

Looking at sample questions is also a good way to identify any areas of weakness in your knowledge. If you find that you struggle with a particular topic, you can then focus your revision on that area.

myCBSEguide offers ample Class 11 Mathematics case study questions, so there is no excuse. With a little bit of preparation, Class 11 Mathematics students can boost their chances of getting the grade they deserve.

Some samples of Class 11 Mathematics case study questions are as follows:

Class 11 Mathematics case study question 1

• 9 km and 13 km
• 9.8 km and 13.8 km
• 9.5 km and 13.5 km
• 10 km and 14 km
• x  ≤   −1913
• x <  −1613
• −1613  < x <  −1913
• There are no solution.
• y  ≤   12 x+2
• y >  12 x+2
• y  ≥   12 x+2
• y <  12 x+2

Answer Key:

• (b) 9.8 km and 13.8 km
• (a) −1913   ≤  x 
• (b)  y >  12 x+2
• (d) (-5, 5)

Class 11 Mathematics case study question 2

• 2 C 1 × 13 C 10
• 2 C 1 × 10 C 13
• 1 C 2 × 13 C 10
• 2 C 10 × 13 C 10
• 6 C 2​ × 3 C 4   × 11 C 5 ​
• 6 C 2​ × 3 C 4   × 11 C 5
• 6 C 2​ × 3 C 5 × 11 C 4 ​
• 6 C 2 ​  ×   3 C 1 ​  × 11 C 5 ​
• (b) (13) 4  ways
• (c) 2860 ways.

Class 11 Mathematics case study question 3

Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the rate of Rs 1000 /yard². The builder planned ground floor of 5 m height, first floor of 4.75 m and so on each floor is 0.25 m less than its previous floor.

Class 11 Mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

• 5x + 6y = 60
• 6x + 5y = 60
• (a) (10, 0)
• (b) 6x + 5y = 60
• (b) 60/√ 61 km
• (d) 2√61 km

A peek at the Class 11 Mathematics curriculum

The Mathematics Syllabus has evolved over time in response to the subject’s expansion and developing societal requirements. The Senior Secondary stage serves as a springboard for students to pursue higher academic education in Mathematics or professional subjects such as Engineering, Physical and Biological Science, Commerce, or Computer Applications. The current updated curriculum has been prepared in compliance with the National Curriculum Framework 2005 and the instructions provided by the Focus Group on Teaching Mathematics 2005 in order to satisfy the rising demands of all student groups. Greater focus has been placed on the application of various principles by motivating the themes from real-life events and other subject areas.

Class 11 Mathematics (Code No. 041)

Design of Class 11 Mathematics exam paper

CBSE Class 11 mathematics question paper is designed to assess students’ understanding of the subject’s essential concepts. Class 11 mathematics question paper will assess their problem-solving and analytical abilities. Before beginning their test preparations, students in Class 11 maths should properly review the question paper format. This will assist Class 11 mathematics students in better understanding the paper and achieving optimum scores. Refer to the Class 11 Mathematics question paper design provided.

 Class 11 Mathematics Question Paper Design

• No chapter-wise weightage. Care to be taken to cover all the chapters.
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.  

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections.

  Prescribed Books:

• Mathematics Textbook for Class XI, NCERT Publications
• Mathematics Exemplar Problem for Class XI, Published by NCERT
• Mathematics Lab Manual class XI, published by NCERT

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Chapter 8 Class 11 Sequences and Series

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Updated for new NCERT - 2023-2024 Edition.

Solutions of Chapter 8 Sequences and Series of Class 11 NCERT book available free. All exercise questions, examples, miscellaneous are done step by step with detailed explanation for your understanding. 

In this Chapter we learn about Sequences

Sequence is any group of numbers with some pattern.

Like 2, 4, 8, 16, 32, 64, 128, 256, ....

1, 2, 3, 4, 5, 6, 7, 8

In this chapter we learn

• What a sequence is - and what is finite, infinite sequence, terms of a sequence
• What a series is - it is the sum of a sequence
• Denoting Sum by sigma Σ , and what it means
• What is an AP (Arithmetic Progression) is, and finding its n th term and sum
• Inserting AP between two numbers
• What is Arithmetic Mean ( AM ), and how to find it
• What is a GP (Geometric Progression) is, and finding its n th term and sum
• Inserting GP between two numbers
• What is Geometric Mean ( GM ), and how to find it
• Relationship between AM and GM
• Sum of special series
• Finding sum of series when n th term is given
• Finding sum of series when n th term is not given

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CBSE Class 11 Maths – Chapter 9 Sequences and Series- Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Sequence and Series : Notes and Study Materials -pdf

• Concepts of  Sequence and Series
• Sequence and Series Master File
• Sequence and Series Revision Notes
• R D Sharma Solution of AP
• R D Sharma Solution of GP
• R D Sharma Solution of Special Series
• NCERT Solution  Sequence and Series
• NCERT  Exemplar Solution Sequence and Series
• Sequence and Series : Solved Example 1

It means an arrangement of number in definite order according to some rule

Important point

• The various numbers occurring in a sequence are called its terms. The number of terms is called the length of the series
• Terms of a sequence are denoted generally by $a_1 , a_2, a_3, ….., a_n$ etc., The subscripts denote the position of the term. We can choose any other letter to denote it
• The nth term is the number at the nth position of the sequence and is denoted by a n
• The nth term is also called the general term of the sequence
• A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it of the type {1, 2, 3…k}.
• Many times is possible to express general term in terms of algebraic formula. But it may not be true in other cases. But we should be able to generate the terms of the sequence using some rules or theoretical scheme

Finite sequence A sequence containing a finite number of terms is called Finite sequence infinite sequence A sequence is called infinite if it is not a finite sequence.

Let a 1 , a 2 , a 3 ..be the sequence, then the sum expressed a 1 + a 2 +a 3 + ……. is called series.

• A series is called finite series if it has got finite number of terms
• A series is called infinite series if it has got infinite terms
• Series are often represented in compact form, called sigma notation, using the Greek letter σ (sigma)
• so , a 1 + a 2 +a 3 + …….a n can be expressed as $\sum_{k=1}^{n}a_k$

Example $1+2+3 + 4+ 5+….+100$ $= \sum_{k=1}^{n}k$

Types of Sequences

Arithmetic progression.

An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant Examples 1. $1,5,9,13,17….$ 2. $1,2,3,4,5,…..$ Important Notes about Arithmetic Progression 1.The difference between any successive members is a constant and it is called the common difference of AP 2. If a 1 , a 2 ,a 3 ,a 4 ,a 5 are the terms in AP then D=a 2 -a 1 =a 3 – a 2 =a 4 – a 3 =a 5 –a­ 4 3. We can represent the general form of AP in the form a,a+d,a+2d,a+3d,a+4d.. Where a is first term and d is the common difference 4. Nth term of Arithmetic Progression is given by n th term = a + (n – 1)d 5. Sum of nth item in Arithmetic Progression is given by $S_n =\frac {n}{2}[a + (n-1)d]$ Or $S_n =\frac {n}{2}[t_1+ t_n]$ Arithmetic Mean The arithmetic mean A of any two numbers a and b is given by (a+b)/2 i.e. a, A, b are in AP A -a = b- A or $A = \frac {(a+b)}{2}$ If we want to add n terms between A and B so that result sequence is in AP Let A 1 , A 2 , A 3 , A 4 , A 5 …. A n be the terms added between a and b Then b= a+ [(n+2) -1]d or $d = \frac {(b-a)}{(n+1)}$ So terms will be $a+ \frac {(b-a)}{(n+1)}, a + 2 \frac {(b-a)}{(n+1)} ,……. a+ n \frac {(b-a)}{(n+1)}$

Geometric Progressions

A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout. Examples 1. $2,4,8,16,32….$ 2. $3,6,12,24,48,…..$ Important Notes about Geometric Progressions

• The ratio of any term to its preceding term is same throughout. This constant factor is called the common ratio.
• If a 1 , a 2 ,a 3 ,a 4 ,a 5 are the terms in AP then $r=\frac {a_2}{a_1} =\frac {a_3}{a_2} =\frac {a_4}{a_3}=\frac {a_5}{a_4}$
• We can represent the general form of G.P in the form a,ar,ar 2 ,ar 3 …… Where a is first term and r is the common ratio
• Nth term of Geometric Progression is given by n th term = ar n-1
• Sum of nth item inGeometric Progression is given by $S_n =a \frac {r^n -1}{r-1}$ Or $S_n =a \frac {1-r^n }{ 1-r}$

Geometric Means The Geometric mean G of any two numbers a and b is given by (ab) 1/2 i.e. a, G, b are in G.P $\frac {G}{a} = \frac {b}{G}$ or $G=\sqrt {ab}$ If we want to add n terms between A and B so that result sequence is in GP Let G 1 , G 2 , G 3 , G 4 , G 5 …. G n be the terms added between a and b Then $b= ar^{n+1}$ or $r= (\frac {b}{a})^{\frac {1}{n+1}}$ So terms will be $a(\frac {b}{a})^{\frac {1}{n+1}}, a(\frac {b}{a})^{\frac {2}{n+1}} ,……. a(\frac {b}{a})^{\frac {n}{n+1}}$ Relationship Between A.P and G.P Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively. Then A ≥ G And $A – G = \frac {{\sqrt {a} – \sqrt {b}}^2}{2}$

Sum of Special Series

Sum of first n natural numbers $1 + 2 + 3 +….. + n = \frac {n(n+1}{2}$ $ \sum_{k=1}^{n} n =\frac {n(n+1}{2}$ sum of squares of the first n natural numbers $1^2+ 2^2+3^2 +…. + n^2 = \frac {n(n+1)(2n+1)}{6}$ $ \sum_{k=1}^{n} n^2= \frac {n(n+1)(2n+1)}{6}$ Sum of cubes of the first n natural numbers $1^3+ 2^3+3^3 +….. + n^3 = \frac {(n(n+1))^2}{4}$ $ \sum_{k=1}^{n} n^3= \frac {(n(n+1))^2}{4}$

Rules for finding the sum of Series

a. Write the nth term T n of the series b. Write the T n in the polynomial form of n $T_n= a n^3 + bn^2 + cn +d$ c. The sum of series can be written as $ \sum S = a \sum {n^3} + b \sum {n^2} + c \sum n + nd We already know the values if these standard from the formula given above and we can easily find the sum of the series Example Find the sum of the series $2^2 + 4^2 + 6^2 + …..(2n)^2} Solution Let nth term T n of the series Then $T_n = (2n)^2 = 4n^2$ Now $2^2 + 4^2 + 6^2 + …..(2n)^2 = \sum_{k=1}^{n} 4k^2 = 4 sum_{k=1}^{n} k^2 = 4 \frac {n(n+1)(2n+1)}{6}$ $=\frac {2n(n+1)(2n+1)}{3}$

Method of Difference

Many times , nth term of the series can be determined. For example 5 + 11 + 19 + 29 + 41…… If the series is such that difference between successive terms are either in A.P or G.P, then we can the nth term using method of difference $S_n = 5 + 11 + 19 + 29 + … + a_{n-1} + a_n$ or $S_n= 5 + 11 + 19 + … + + a_{n-1} + a_n$ On subtraction, we get $0 = 5 + [6 + 8 + 10 + 12 + …(n – 1) terms] – a_n$ Here 6,8,10 is in A.P,So $a_n = 5 + \frac {n-1}{2} [12 + (n-1)2]$ or $ a_n= n^2 + 3n + 1$ Now it is easy to find the Sum of the series $S_n = \sum_{k=1}^{n} {k^2 +3k +1}$ $=\frac {n(n+2)(n+4)}{3}$

Sequences and Series Class 11 MCQs Questions with Answers

Question 1. If a, b, c are in G.P., then the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root if d/a, e/b, f/c are in (a) AP (b) GP (c) HP (d) none of these

Answer: (a) AP Hint: Given a, b, c are in GP ⇒ b² = ac ⇒ b² – ac = 0 So, ax² + 2bx + c = 0 have equal roots. Now D = 4b² – 4ac and the root is -2b/2a = -b/a So -b/a is the common root. Now, dx² + 2ex + f = 0 ⇒ d(-b/a)² + 2e×(-b/a) + f = 0 ⇒ db2 /a² – 2be/a + f = 0 ⇒ d×ac /a² – 2be/a + f = 0 ⇒ dc/a – 2be/a + f = 0 ⇒ d/a – 2be/ac + f/c = 0 ⇒ d/a + f/c = 2be/ac ⇒ d/a + f/c = 2be/b² ⇒ d/a + f/c = 2e/b ⇒ d/a, e/b, f/c are in AP

Question 2. If a, b, c are in AP then (a) b = a + c (b) 2b = a + c (c) b² = a + c (d) 2b² = a + c

Answer: (b) 2b = a + c Hint: Given, a, b, c are in AP ⇒ b – a = c – b ⇒ b + b = a + c ⇒ 2b = a + c

Question 3: Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in Ap. The common ratio of GP is (a) 2 + √3 (b) 2 – √3 (c) 2 ± √3 (d) None of these

Answer: (a) 2 + √3 Hint: Let the three numbers be a/r, a, ar Since the numbers form an increasing GP, So r > 1 Now, it is given that a/r, 2a, ar are in AP ⇒ 4a = a/r + ar ⇒ r² – 4r + 1 = 0 ⇒ r = 2 ± √3 ⇒ r = 2 + √3 {Since r > 1}

Question 4: The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is (a) n/(n+1) (b) 1/(n+1) (c) 1/n (d) None of these

Answer: (a) n/(n+1) Hint: Given series is: S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1) ⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1)) ⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1) ⇒ S = 1 – 1/(n+1) ⇒ S = (n + 1 – 1)/(n+1) ⇒ S = n/(n+1)

Question 5: If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then (a) a, b, c are in AP (b) a², b², c² are in AP (c) 1/1, 1/b, 1/c are in AP (d) None of these

Answer: (b) a², b², c² are in AP Hint: Given, 1/(b + c), 1/(c + a), 1/(a + b) ⇒ 2/(c + a) = 1/(b + c) + 1/(a + b) ⇒ 2b² = a² + c² ⇒ a², b², c² are in AP

Question 6: The sum of series 1/2! + 1/4! + 1/6! + ….. is (a) e² – 1 / 2 (b) (e – 1)² /2 e (c) e² – 1 / 2 e (d) e² – 2 / e

Answer: (b) (e – 1)² /2 e Hint: We know that, e x = 1 + x/1! + x² /2! + x³ /3! + x 4 /4! + ……….. Now, e 1 = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ……….. e -1 = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ……….. e 1 + e -1 = 2(1 + 1/2! + 1/4! + ………..) ⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..) ⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..) ⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ……….. ⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ……….. ⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ……….. ⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..

Question 7: The third term of a geometric progression is 4. The product of the first five terms is (a) 4 3 (b) 4 5 (c) 4 4 (d) none of these

Answer: (b) 4 5 Hint: here it is given that T 3 = 4. ⇒ ar² = 4 Now product of first five terms = a.ar.ar².ar³.ar 4 = a 5 r 10 = (ar 2 ) 5 = 4 5

Question 8: Let Tr be the r th term of an A.P., for r = 1, 2, 3, … If for some positive integers m, n, we have Tm = 1/n and Tn = 1/m, then Tm n equals (a) 1/m n (b) 1/m + 1/n (c) 1 (d) 0

Answer: (c) 1 Hint: Let first term is a and the common difference is d of the AP Now, T m = 1/n ⇒ a + (m-1)d = 1/n ………… 1 and T n = 1/m ⇒ a + (n-1)d = 1/m ………. 2 From equation 2 – 1, we get (m-1)d – (n-1)d = 1/n – 1/m ⇒ (m-n)d = (m-n)/mn ⇒ d = 1/mn From equation 1, we get a + (m-1)/mn = 1/n ⇒ a = 1/n – (m-1)/mn ⇒ a = {m – (m-1)}/mn ⇒ a = {m – m + 1)}/mn ⇒ a = 1/mn Now, T mn = 1/mn + (mn-1)/mn ⇒ T mn = 1/mn + 1 – 1/mn ⇒ T mn = 1

Question 9. The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is (a) 2 (b) 4 (c) 6 (d) 8

Answer: (c) 6 Hint: Let a and b are two numbers such that a + b = 13/6 Let A 1 , A 2 , A 3 , ………A 2n be 2n arithmetic means between a and b Then, A 1 + A 2 + A 3 + ………+ A 2n = 2n{(n + 1)/2} ⇒ n(a + b) = 13n/6 Given that A 1 + A 2 + A 3 + ………+ A 2n = 2n + 1 ⇒ 13n/6 = 2n + 1 ⇒ n = 6

Question 10. If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in (a) A.P. (b) G.P. (c) H.P. (d) A.G.P.

Answer: (c) H.P. Hint: Given, equation is ax² + bx + c = 0 Let p and q are the roots of this equation. Now p+q = -b/a and pq = c/a Given that p + q = 1/p² + 1/q² ⇒ p + q = (p² + q²)/(p² ×q²) ⇒ p + q = {(p + q)² – 2pq}/(pq)² ⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)² ⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a} ⇒ -bc²/a³ = {b² – 2ca}/a² ⇒ -bc²/a = b² – 2ca Divide by bc on both side, we get ⇒ -c /a = b/c – 2a/b ⇒ 2a/b = b/c + c/a ⇒ b/c, a/b, c/a are in AP ⇒ c/a, a/b, b/c are in AP ⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP ⇒ a/c, b/a, c/b are in HP

Question 11. If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then (a) a, b, c are in AP (b) a², b², c² are in AP (c) 1/1, 1/b, 1/c are in AP (d) None of these

Question 12. The 35th partial sum of the arithmetic sequence with terms a n = n/2 + 1 (a) 240 (b) 280 (c) 330 (d) 350

Answer: (d) 350 Hint: The 35th partial sum of this sequence is the sum of the first thirty-five terms. The first few terms of the sequence are: a 1 = 1/2 + 1 = 3/2 a 2 = 2/2 + 1 = 2 a 3 = 3/2 + 1 = 5/2 Here common difference d = 2 – 3/2 = 1/2 Now, a 35 = a 1 + (35 – 1)d = 3/2 + 34 ×(1/2) = 17/2 Now, the sum = (35/2) × (3/2 + 37/2) = (35/2) × (40/2) = (35/2) × 20 = 35 × 10 = 350

Question 13. The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is (a) 2 (b) 4 (c) 6 (d) 8

Question 14. The first term of a GP is 1. The sum of the third term and fifth term is 90. The common ratio of GP is (a) 1 (b) 2 (c) 3 (d) 4

Answer: (c) 3 Hint: Let first term of the GP is a and common ratio is r. 3rd term = ar² 5th term = ar 4 Now ⇒ ar² + ar 4 = 90 ⇒ a(r² + r 4 ) = 90 ⇒ r² + r 4 = 90 ⇒ r² ×(r² + 1) = 90 ⇒ r²(r² + 1) = 3² ×(3² + 1) ⇒ r = 3 So the common ratio is 3

Question 15. The sum of AP 2, 5, 8, …..up to 50 terms is (a) 3557 (b) 3775 (c) 3757 (d) 3575

Answer: (b) 3775 Hint: Given, AP is 2, 5, 8, …..up to 50 Now, first term a = 2 common difference d = 5 – 2 = 3 Number of terms = 50 Now, Sum = (n/2)×{2a + (n – 1)d} = (50/2)×{2×2 + (50 – 1)3} = 25×{4 + 49×3} = 25×(4 + 147) = 25 × 151 = 3775

Question 16. If 2/3, k, 5/8 are in AP then the value of k is (a) 31/24 (b) 31/48 (c) 24/31 (d) 48/31

Answer: (b) 31/48 Hint: Given, 2/3, k, 5/8 are in AP ⇒ 2k = 2/3 + 5/8 ⇒ 2k = 31/24 ⇒ k = 31/48 So, the value of k is 31/48

Question 17. The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is (a) n/(n+1) (b) 1/(n+1) (c) 1/n (d) None of these

Answer: (a) n/(n+1) Hint: Given series is: S = (1/1·2) + (1/2·3) + (1/3·4) – ……………….1/n.(n+1) ⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -………(1/n – 1/(n+1)) ⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1) ⇒ S = 1 – 1/(n+1) ⇒ S = (n + 1 – 1)/(n+1) ⇒ S = n/(n+1)

Question 18. If the third term of an A.P. is 7 and its 7 th term is 2 more than three times of its third term, then the sum of its first 20 terms is (a) 228 (b) 74 (c) 740 (d) 1090

Answer: (c) 740 Hint: Let a is the first term and d is the common difference of AP Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term ⇒ a + 2d = 7 ………….. 1 and 3(a + 2d) + 2 = a + 6d ⇒ 3×7 + 2 = a + 6d ⇒ 21 + 2 = a + 6d ⇒ a + 6d = 23 ………….. 2 From equation 1 – 2, we get 4d = 16 ⇒ d = 16/4 ⇒ d = 4 From equation 1, we get a + 2×4 = 7 ⇒ a + 8 = 7 ⇒ a = -1 Now, the sum of its first 20 terms = (20/2)×{2×(-1) + (20-1)×4} = 10×{-2 + 19×4)} = 10×{-2 + 76)} = 10 × 74 = 740

Question 19. If the sum of the first 2n terms of the A.P. 2, 5, 8, ….., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ….., then n equals (a) 10 (b) 12 (c) 11 (d) 13

Answer: (c) 11 Hint: Given, the sum of the first 2n terms of the A.P. 2, 5, 8, …..= the sum of the first n terms of the A.P. 57, 59, 61, …. ⇒ (2n/2)×{2×2 + (2n-1)3} = (n/2)×{2×57 + (n-1)2} ⇒ n×{4 + 6n – 3} = (n/2)×{114 + 2n – 2} ⇒ 6n + 1 = {2n + 112}/2 ⇒ 6n + 1 = n + 56 ⇒ 6n – n = 56 – 1 ⇒ 5n = 55 ⇒ n = 55/5 ⇒ n = 11

Question 20. If a is the A.M. of b and c and G 1 and G 2 are two GM between them then the sum of their cubes is (a) abc (b) 2abc (c) 3abc (d) 4abc

Answer: (b) 2abc Hint: Given, a is the A.M. of b and c ⇒ a = (b + c) ⇒ 2a = b + c ………… 1 Again, given G 1 and G 1 are two GM between b and c, ⇒ b, G 1 , G 2 , c are in the GP having common ration r, then ⇒ r = (c/b) 1/(2+1) = (c/b) 1/3 Now, G 1 = br = b×(c/b) 1/3 and G 1 = br = b×(c/b) 2/3 Now, (G 1 )³ + (G 2 )3 = b³ ×(c/b) + b³ ×(c/b)² ⇒ (G 1 )³ + (G 2 )³ = b³ ×(c/b)×( 1 + c/b) ⇒ (G 1 )³ + (G 2 )³ = b³ ×(c/b)×( b + c)/b ⇒ (G 1 )³ + (G 2 )³ = b² ×c×( b + c)/b ⇒ (G 1 )³ + (G 2 )³ = b² ×c×( b + c)/b ………….. 2 From equation 1 2a = b + c ⇒ 2a/b = (b + c)/b Put value of(b + c)/b in eqaution 2, we get (G 1 )³ + (G 2 )³ = b² × c × (2a/b) ⇒ (G 1 )³ + (G 2 )³ = b × c × 2a ⇒ (G 1 )³ + (G 2 )³ = 2abc

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• Sequences and Series Class 11 Notes CBSE Maths Chapter 9 (Free PDF Download)
• Revision Notes

Revision Notes for CBSE Class 11 Maths Chapter 9 (Sequences and Series) - Free PDF Download

A "sequence" is nothing but an ordered list of numbers. The numbers that are present in the ordered list are called as  "elements" or "terms" of the sequence. When you add up all the terms in a sequence, you get a "series"; the addition, as well as the resulting value, is called the "sum" or "summation." For example, the sequence "1, 2, 3, 4" contains the terms "1", "2", "3", and "4"; the corresponding series is the sum "1 + 2 + 3 + 4", and the series' value is 10. The Sequence and Series Class 11 Notes is one of the important materials when it comes to understanding the basic topics and complex problems in the chapter. With the help of revision notes students can revise the syllabus in a concise manner, right from definitions of sequence, Series and Progressions to important problems from exam point of view. The first chapter includes sequences and series, as well as their key properties. Topics like increasing, decreasing, bounded, convergent, and divergent sequences are discussed at basic level, which is appropriate for a 11-grade student. A.P. and G.P. are explained in detail and important problems are addressed and solved. The famous Fibonacci type sequences are demonstrated, as well as different methods for finding formulae for the nth term of a recursive sequence and recursive formulas for other known series. You'll find new ways to find the nth term and partial sums for non-geometric and non-arithmetic sequences in Class 11 Maths Chapter 9 Notes.

Download CBSE Class 11 Maths Revision Notes 2023-24 PDF

Also, check CBSE Class 11 Maths revision notes for all chapters:

Sequences and Series Chapter-Related Important Study Materials It is a curated compilation of relevant online resources that complement and expand upon the content covered in a specific chapter. Explore these links to access additional readings, explanatory videos, practice exercises, and other valuable materials that enhance your understanding of the chapter's subject matter.

Sequences and Series Class 11 Notes Maths - Basic Subjective Questions

Section–A (1 Mark Questions)

1. If the sum of n terms of an A.P. is given by S n =3n+2n 2 then find the common difference of the A.P.

Ans. Given, $S_n=3 n+2 n^2$

$$ \begin{aligned} & S_1=3(1)+2(1)^2=5=a_1 \\ & S_2=3(2)+2(2)^2=14=a_1+a_2 \\ & \therefore S_2-S_1=9=a_2 \\ & \therefore d=a_2-a_1=9-5=4 . \end{aligned} $$

2. If the third term of G.P. is 4, then what is the product of its first 5 terms.

Ans. Let $a$ and $r$ be the first term and common ratio of G.P., respectively.

Given that the third term is 4 .

$$\therefore a r^2=4$$

Product of first 5 terms

$$=a \cdot a r \cdot a r^2 \cdot a r^3 \cdot a r^4=a^5 r^{10}=\left(a r^2\right)^5=4^5\text {. }$$

3. The 17 th term from the end of A.P. -36,-31,-26,.....79 is ________. 

Ans. Here, $a=-36$ and $d=-31-(-36)=5$

$$\begin{aligned}& l=79 \\& \therefore 17^{\text {th }} \text { term from the end }=l-(n-1) d \\& =79-(17-1)(5)=79-80=-1 .\end{aligned}$$

4. Sum of the series $3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....$ n terms, is ________.

Ans. The formula for the summation of $n$ terms of an geometric series is $S_n=\frac{a\left(1-r^n\right)}{1-r}$, where $a$ is the first term in the series and $r$ is the rate of change between successive terms.

Here $a=3$ and $r=\frac{1}{3}$

$$S_n=\frac{3\left(1-\left(\frac{1}{3}\right)^n\right)}{1-\left(\frac{1}{3}\right)}=\frac{9}{}\left(1-\left(\frac{1}{3}\right)^n\right) \text {. }$$

5. The first two terms of the sequence defined by a 1 =3 and a n =3a n-1 + 2 for all n>1 ________. 

Ans. Given: $a_1=3$ and $a_n=3 a_{n-1}+2$, for all $n>1$

When $n=2$ :

$$\begin{aligned}& a_2=3 a_{2-1}=3 a_1+2=3(3)+2 \\& =9+2=11 .\end{aligned}$$

Section-B (2 Marks Questions)

6. If 9 times the 9th term of an A.P. is equal to 13 times the 13th term, then prove that the 22nd term of the A.P. is zero.

Ans. Let the first term and common difference of given A.P. be $a$ and $d$, respectively.

It is given that $9 a_9=13 a_{13}$

$$ \begin{aligned} & \Rightarrow 9(a+8 d)=13(a+12 d) \\ & \Rightarrow 9 a+72 d=13 a+156 d \\ & \Rightarrow 4 a+84 d=0 \\ & \Rightarrow(a+21 d)=0 \\ & \Rightarrow a_{22}=0 . \end{aligned} $$

7. If a,b and c are in G.P., then find the value of $\frac{a-b}{b-c}$ .

Ans. Given that, $a, b$ and $c$ are in G.P.

$\Rightarrow b=a r$ and $c=a r^2$, where $r$ is the common ratio.

$$\Rightarrow \frac{a-b}{b-c}=\frac{a-a r}{a r-a r^2}=\frac{a(1-r)}{ar(1-r)}=\frac{1}{r}=\frac{a}{b} o r \frac{b}{c}$$

8. The sum of terms equidistant from the beginning and end in an A.P. is equal to ________.

Ans. Let $a$ be the first term and $d$ be the common difference of the A.P.

$a_r=r^{4 t}$ term from the beginning $=a+(r-1) d$

$a_r^{\prime}=r^{\text {di }}$ term from the end $=(a+(n-1) d)+(r-1)(-d)$

(as first term is $a_n=a+(n-1) d$ and common difference is ' $-d^{\prime}$ )

$$a_r+a^{\prime} r=a+(r-1) d+(a+(n-1) d)+(r-1)(-d)$$

$=2 a+(n-1) d$, which is independent of ' $r$ '

Thus, sum of the terms equidistant from the beginning and end in an A.P. is constant.

9. A man saved Rs. 66000 in 20 years. In each succeeding year after the first year, he saved Rs. 200 more than what he saved in the previous year. How much did he save in the first year?

Ans. Let us assume that the man saved Rs. $a$ in the first year.

In each succeeding year, an increment of Rs. 200 is made.

So, it forms an A.P. whose first term $=a$, common difference, $d=200$ and $n=20$

$$ \begin{aligned} & \therefore S_{20}=\frac{20}{2}[2 a+(20-1) d] \\ & \Rightarrow 66000=10[2 a+19 \times 200] \\ & \Rightarrow 6600=2 a+19 \times 200 \\ & \Rightarrow 2 a=2800 \\ & \therefore a=1400 . \end{aligned} $$

10. The sum of interior angles of a triangle is $180^{\circ}$. Show that the sum of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21-sided polygon.

Ans. We know that the sum of interior angles of a polygon of side $n$ is $(n-2) \times 180^{\circ}$.

Let $a_n=(n-2) \times 180^{\circ}$

Since $a_n$ is linear in $n$, it is $n^{\text {,h }}$ term of some A.P.

$$a_3=(3-2) \times 180^{\circ}=180^{\circ}$$

Common difference, $d=180^{\circ}$

Sum of the interior angles for a 21-sided polygon is:

$$a_{21}=(21-2) \times 180^{\circ}=3420^{\circ} \text {. }$$

11. Find the $r^{th}$ term of an A.P., whose sum of first n terms is $2n+3n^{2}$ .

Ans. Sum of $n$ terms of A.P., $S_n=2 n+3 n^2$

$$a_n=S_n-S_{n-1}$$

$$\begin{aligned}& =\left(2 n+3 n^2\right)-\left[2(n-1)+3(n-1)^2\right] \\& =[2 n-2(n-1)]+\left[3 n^2-3(n-1)^2\right] \\& =2(n-n+1)+3(n-n+1)(n+n-1) \\& =2+3(2 n-1) \\& =6 n-1 \\& \therefore r^{\text {sn }} \text { term } a_r=6 r-1 .\end{aligned}$$

12. Show that the products of the corresponding terms of the sequences $a,ar,ar^{2},....ar^{n-1}$ and $A,AR,AR^{2},....AR^{n-1}$ form a G.P., and find the common ratio.

Ans. It has to be proved that the sequence, $a A, \operatorname{ar} A R, a r^2 A R^2, \ldots a r^{n-1} A R^{n-1}$, forms G.P

$\dfrac{\text { Second term }}{\text { First term }}=\dfrac{a r A R}{a A}=r R$

$\dfrac{\text { Third term }}{\text { Second term }}=\dfrac{a r^2 A R^2}{a r A R}=r R$

Thus, the above sequenee forms a G.P. and the common ratio is $r R$.

13. If the sum of n terms of an A.P. is $(pn+qn^{2})$, where p and q are constants, find the common difference.

Ans. It is known that,

$$S_n=\frac{n}{2}[2 a+(n-1) d]$$

According to the given condition,

$$\begin{aligned}& \frac{n}{2}[2 a+(n-1) d]=p n+q n^2 \\& \Rightarrow \frac{n}{2}[2 a+n d-d]=p n+q n^2 \\& \Rightarrow n a+n^2 \frac{d}{2}-n \cdot \frac{d}{2}=p n+qn^2\end{aligned}$$

Comparing the coefficients of $n^2$ on both sides, we obtain

$$\frac{d}{2}=q \Rightarrow d=2 q$$

Thus, the common difference of the A.P. is $2 q$.

PDF Summary - Class 11 Maths Sequences and Series Notes (Chapter 9)

1. Definition:

Any function with domain as a set of natural numbers is called sequence.

Real sequence: Sequence with range as subset of real numbers.

For example:-

If \[{a_1}\;,\;{a_2}\;,{a_3}\;,\;...............\;,\;{a_n}\]is a sequence, then \[{a_1}\; + \;{a_2}\; + \;{a_3}\; + \;...............\; + \;{a_n}\]  is a series.

Progression: When terms of a sequence follow a certain pattern. 

But it is not always necessary that terms of sequence follow a certain pattern.

1.1 Arithmetic Progression (AP):

An arithmetic progression is a sequence of numbers in which each successive term is a sum of its preceding term and a fixed number.

If this fixed number is positive, then it is an increasing AP and if this fixed number is negative, then it is a decreasing AP.

This fixed term is called common difference and is usually represented by ‘d’.

Let ‘a’ be the first term of an AP.

N th term of an AP: \[{t_n}\; = \;a\; + \left( {n\; - \;1} \right)d\;,\;{\text{where}}\;\;d\; = \;{a_n} - \;{a_{n - 1}}\]

Sum of first N terms of an AP: \[{S_n}\; = \;\dfrac{n}{2}\left[ {a\; + \;\left( {n\; - \;1} \right)d} \right]\; = \;\dfrac{n}{2}\left[ {a\; + \;l} \right]\;{\text{where}}\;{\text{,}}\;l\;{\text{is}}\;{\text{last}}\;{\text{term}}\;{\text{of}}\;{\text{an}}\;{\text{AP}}\].

Properties of an AP:

Increasing, Decreasing, Multiplying and dividing each term of an AP by a non-zero constant results into an AP.

3 numbers in an AP: \[a\; - \;d\;,\;a\;,\;a\; + \;d\]

4 numbers in an AP: \[a\; - \;3d\;,\;a\; - \;d\;,\;a\; + \;d\;,\;a\; + \;3d\]

5 numbers in an AP: \[a\; - \;2d\;,\;a\; - \;d\;,\;a\;,\;a\; + \;d\;,\;a\; + \;2d\]

6 numbers in an AP: \[a\; - \;5d\;,\;a\; - \;3d\;,\;a\; - \;d\;,\;a\; + \;d\;,\;a\; + \;3d\;,\;a\; + \;5d\]

An AP can have zero, positive or negative common difference.

 The sum of the two terms of an AP equidistant from the beginning & end is constant and equal to the sum of first & last terms.

Any term of an AP (except the first) is equal to half the sum of terms which are equidistant from it.

\[ \Rightarrow \;\;{a_n}\; = \;\dfrac{1}{2}\left( {{a_{n\; - \;k}}\; + \;{a_{n\; + \;k}}} \right)\;\;,\;k\; < \;n\]

\[{t_r}\; = \;{S_r}\; - \;{S_{r\; - \;1}}\]

If three numbers are in AP : a, b, c are in AP \[ \Rightarrow \;2b\; = \;a\; + \;c\]

N th term of an AP is a linear expression in n: \[An\; + \;B\]where A is the common difference of an AP.

1.2 Geometric Progression (GP):

It is a sequence in which each term is obtained by multiplying the preceding term by a fixed number (which is constant) called common ratio. First term of GP is non zero.

Common ratio can be obtained by dividing a term by its consecutive preceding term.

If ‘a’ is the first term and ‘r’ is the common ratio then,

GP is \[a\;,ar\;,\;a{r^2}\;,\;a{r^3}\;,\;a{r^4}\;,\;.\;.\;.\;.\;.\;.\;.\;\]

N th term of a GP: \[{t_n}\; = \;a{r^{n - 1}}\]

Sum of first N terms of a GP: \[{s_n}\; = \;\dfrac{{a\left( {1\; - \;{r^n}} \right)}}{{\left( {1\; - \;r} \right)}}\;,\;r\; \ne \;1\]

Sum of infinite GP when \[|r|\; < \;1\;\;\& \;\;n\; \to \;\infty \]

\[|r|\; < \;1\;\; \Rightarrow \;{r^n}\; \to \;0\; \Rightarrow \;{S_\infty }\; = \;\dfrac{a}{{1\; - \;r}}\]

Properties of a GP:

Multiplying and dividing each term of a GP by a non- zero constant results into a GP.

Reciprocal of terms of GP is also GP.

3 consecutive terms in GP: \[\dfrac{a}{r}\;,\;a\;,\;ar\]

4 consecutive terms in GP: \[\dfrac{a}{{{r^2}}}\;,\;\dfrac{a}{r}\;,\;ar\;,\;a{r^2}\]

If three numbers are in GP : a, b, c are in GP \[ \Rightarrow \;{b^2}\; = \;ac\]

Each term of a GP raised to the same power also forms a G.P.

Choosing terms of GP at regular intervals also forms a GP.

The product of the terms equidistant from the beginning and the last is always same and is equal to the product of the first and the last term for a finite GP.

If \[{a_1}\;,\;{a_2}\;,{a_3}\;,\;...............\;,\;{a_n}\] forms GP with non-zero and non-negative terms then\[\log {a_1}\;,\;\log {a_2}\;\log ,{a_3}\;,\;...............\;,\;\log {a_n}\] are in GP or vice versa.

2.1 Arithmetic Mean

When three terms are in AP, the middle term is called AM between the other two.

If a, b, c are in AP, b is AM between a and c.

If n positive terms \[{a_1}\;,\;{a_2}\;,{a_3}\;,\;...............\;,\;{a_n}\] are in AP, then AM is:

\[A\; = \;\dfrac{{{a_1}\; + \;{a_2}\; + \;{a_3}\; + \;.......\; + \;{a_n}}}{n}\]

2.2 n-Arithmetic Means Between Two Numbers

If a, b are two numbers and \[a\;,\;{a_1}\;,\;{a_2}\;,{a_3}\;,\;...............\;,\;{a_n}\;,\;b\]are in an AP, then

\[{a_1}\;,\;{a_2}\;,{a_3}\;,\;...............\;,\;{a_n}\] are n AM’s between a and b.

\[{A_1}\; = \;a\; + \;d\;,\;{A_2}\; = \;a\; + \;2d\;,\;...........\;,\;{A_n}\; = \;a\; + \;nd\], where \[d\; = \;\dfrac{{b\; - \;a}}{{n\; + \;1}}\]

NOTE: Sum of n AM’s inserted between a and b is equal to n times a single AM between a and b.\[\Rightarrow \; \sum\limits_{r\; = \;1}^n {{A_r}}\; = \;nA\]

2.3 Geometric Mean

If \[a,{\text{ }}b,{\text{ }}c\]are in GP, then b is called GM between a and c.

So, \[{b^2}\; = \;ac\;\;or\;\;b\; = \;\sqrt {ac} \;;\;a > 0\;,\;b > 0\]

2.4 n-Geometric Means between two numbers 

If a, b are two numbers and \[a\;,\;{G_1}\;,\;{G_2}\;,{G_3}\;,\;...............\;,\;{G_n}\;,\;b\]are in a GP, then

\[{G_1}\;,\;{G_2}\;,{G_3}\;,\;...............\;,\;{G_n}\] are n GM’s between a and b.

\[{G_1}\; = \;ar\;,\;{G_2}\; = \;a{r^2}\;,..........,\;{G_n}\; = \;a{r^{n\; - \;1}}\], where \[r\; = \;{\left( {\dfrac{b}{a}} \right)^{\dfrac{1}{{n\; + \;1}}}}\]

NOTE: Product of n GM’s inserted between a and b is equal to n th power of single GM between a and b. \[ \Rightarrow \;{\prod\limits_{r\; = \;1}^n {{G_r}\; = \;\left( G \right)} ^n}\]

2.5 Arithmetic, Geometric and Harmonic means between two given numbers

Let A, G and H be the arithmetic, geometric and harmonic mean between two integers numbers a and b.

\[ \Rightarrow \;A\; = \;\dfrac{{a\; + \;b}}{2}\;,\;G\; = \;\sqrt {ab} \;,\;H\; = \;\dfrac{{2ab}}{{a\; + \;b}}\]

The three means have following three properties:

\[A\; \geqslant \;G\; \geqslant \;H\]

\[{G^2}\; = \;AH\] which means that A, G, H forms a GP.

Equation \[{x^2}\; - \;2Ax\; + \;{G^2}\; = \;0\] have a and b as its roots.

If A, G, H are corresponding means between three given numbers a, b and c, then the equation having a, b, c as its roots is \[{x^3}\; - \;3A{x^2}\; + \;\dfrac{{3{G^2}}}{H}x\; - \;{G^3}\; = \;0\]

NOTE:  Some important properties of Arithmetic & Geometric Means between two quantities:

If A and G are arithmetic and geometric mean between a and b then Quadratic equation \[{x^2}\; - \;2Ax\; + \;{G^2}\; = \;0\]has a and b as its roots.

If A and G are AM and GM between two numbers a and b, then

\[a\; = \;A\; + \;\sqrt {{A^2}\; - \;{G^2}} \], \[b\; = \;A\; - \;\sqrt {{A^2}\; - \;{G^2}} \]

3. Sigma Notations:

3.1 Theorems

(i) \[\sum\limits_{r\; = \;1}^n {({a_r}\; + \;{b_r})\; = \;\sum\limits_{r\; = \;1}^n {{a_r}\; + \;\sum\limits_{r\; = \;1}^n {{b_r}} } } \]

(ii) \[\sum\limits_{r\, = \;1}^n {ka\; = \;k\;\sum\limits_{r\, = \;1}^n {{a_r}} } \]

(iii) \[\sum\limits_{r\, = \;1}^n k \; = \;nk\]

4. Sum of n Terms of Some Special Sequences

4.1 Sum of first n natural numbers

\[\sum\limits_{k\, = \;1}^n k \; = \;1\; + \;2\; + \;3\; + \;.......\; + \;n\; = \;\dfrac{{n\left( {n\; + \;1} \right)}}{2}\]

4.2 Sum of squares of  first n natural numbers

\[\sum\limits_{k\, = \;1}^n {{k^2}} \; = \;{1^2}\; + \;{2^2}\; + \;{3^2}\; + \;.......\; + \;{n^2}\; = \;\dfrac{{n\left( {n\; + \;1} \right)\left( {2n\; + \;1} \right)}}{6}\]

4.3 Sum of cubes of  first n natural numbers 

\[\sum\limits_{k\, = \;1}^n {{k^3}} \; = \;{1^3}\; + \;{2^3}\; + \;{3^3}\; + \;.......\; + \;{n^3}\; = \;{\left[ {\dfrac{{n\left( {n\; + \;1} \right)}}{2}} \right]^2}\; = \;{\left[ {\sum\limits_{k\, = \;1}^n k } \right]^2}\]

5. Arithmetico-Geometric series

An arithmetic-geometric progression (A.G.P.) is a progression in which each term can be represented as the product of the terms of an arithmetic progression (AP) and a geometric progression (GP).

\[AP\;:\;1\;,\;3\;,\;5\;,\;..........\] and \[GP\;;\;1\;,\;x\;,\;{x^2}\;,........\]

\[ \Rightarrow \;\;AGP\;:\;1\;,\;3x\;,\;5{x^2},........\]

5.1 Sum of n terms of an Arithmetico-Geometric Series

${{\text{S}}_n} = {\text{a}} + ({\text{a}} + {\text{d}}){\text{r}} + ({\text{a}} + 2\;{\text{d}}){{\text{r}}^2} +  \ldots  \ldots  + $ $[a + (n - 1)d]{r^{n - 1}}$

then ${S_n} = \dfrac{a}{{1 - r}} + \dfrac{{dr\left( {1 - {r^{n - 1}}} \right)}}{{{{(1 - r)}^2}}} - \dfrac{{[a + (n - 1)d]{r^n}}}{{1 - r}},r \ne 1$

5.2 Sum to Infinity

If $|r| < 1{\text{ \& }}\;n \to \infty $, then $\mathop {\lim }\limits_{n \to \infty }  = 0.\;{S_\infty } = \dfrac{a}{{1 - r}} + \dfrac{{dr}}{{{{(1 - r)}^2}}}$.

6. Harmonic Progression (HP)  

A sequence, reciprocal of whose terms forms an AP is called HP.

If the sequence ${a_1},{a_2},{a_3}, \ldots  \ldots  \ldots  \ldots  \ldots ,{a_n}$ is an HP, then

$\dfrac{1}{{{a_1}}}\;,\;\dfrac{1}{{{a_2}}}\;,\;\dfrac{1}{{{a_3}}}\;,\;.......\;,\;\dfrac{1}{{{a_n}}}$ is an AP or vice versa. There is no formula for the sum of the $n$ terms of an HP. For HP with first terms is a and second term is $b$, then ${n^{{\text{th }}}}$ term is ${t_n} = \dfrac{{ab}}{{b + (n - 1)(a - b)}}$

If \[a,\;b,\;c\] are in ${\text{HP}} \Rightarrow {\text{b}} = \dfrac{{2{\text{ac}}}}{{{\text{a}} + {\text{c}}}}$ or $\dfrac{{\text{a}}}{{\text{c}}} = \dfrac{{{\text{a}} - {\text{b}}}}{{{\text{b}} - {\text{c}}}}$.

7. Harmonic Mean

If \[a,\;b,\;c\] are in HP then, \[b\] is the HM between \[a\;\& \;c\] \[ \Rightarrow \;b\; = \;\dfrac{{2ac}}{{a\; + \;c}}\]. 

Sequence and Series Class 11 Notes

Preparing from CBSE Sequence and Series Notes helps students to understand the important topics such as A.P, G.P, harmonic progressions, the arithmetic-geometric mean, and harmonic mean. These notes help students to get a good score in examinations. Topics are explained in very easy language which helps the students to understand and revise syllabus with almost no time in  Sequences and Series Revision Notes. Students can solve any MCQs and Subjective question paper, once they are thorough with the notes. So students are advised to study Class 11 Maths Chapter 9 Notes without any confusion. Let’s look at the topics covered in these notes. 

Meaning of Sequence

What is a sequence in Math?

Finite Sequence

Infinite Sequence

Types of Sequence

Arithmetic sequence, geometric sequence.

Fibonacci Sequence

Meaning of Series

Notation of Series

Finite and Infinite Series

Types of Series

Arithmetic Series

Geometric Series

Meaning of Geometric Progression (G.P.)

Meaning of Arithmetic Progression (A.P.)

Arithmetic Mean

Geometric Mean

Relation between A.M. and G.M.

Special Series

Sum to n terms of Special Series

A sequence is nothing but a group of objects that follow some particular pattern. If we have some objects which are listed in some kind order so that it has 1st term, 2nd term and so on, then it is a sequence.

What is a Sequence in Math?

In Mathematics, it is defined as a group of numbers which are in an ordered form which follows a particular pattern is called Sequence. There are Finite Sequences and Infinite Sequences. The sequence which has a finite number of terms (Limited terms) is called Finite Sequence. The sequence that has an unlimited number of terms (Infinite terms) is called Infinite Sequence.

There are 3 types of sequences:

In any sequence, if the difference between every successive term is a constant then it is defined as Arithmetic Sequence. It can be in ascending or descending order, but it has to be according to a constant number.

In any sequence, if the ratio between each successive term is constant then it is known as Geometric Sequence. It can be in ascending or descending order according to the constant ratio.

Like we discussed a few topics above, Class 12 Maths, Chapter 9 Sequences and Series is a difficult subject with many problems and concepts. Many of the definitions are thoroughly clarified. As a result, learning all of these will require some extra effort, and students will need to keep revising and practising in order to completely master the subject. Though students may not have enough time to prepare notes on their own, we at Vedantu provide well-organized CBSE Class 11 Maths Notes Chapter 9 Sequences and Series that will assist them in their examination preparation as well as increase their interest in the concepts. Refer to the free PDF of CBSE Sequence and Series Notes for the complete notes.

Tips to Prepare for Exams Using CBSE Sequence and Series Notes

You must complete the previous year's questions after you have completed the concepts and numerical. With the previous year's problems, you'll be able to see just where you're missing and how to progress accordingly.

To improve your pace and accuracy, take online mock tests on a regular basis. This activity will be particularly beneficial in JEE Mains.

Understand your strengths and weaknesses, and work to improve both.

If you notice any questions that seem to be critical when practicing, make a note of them. You must solve the question again later when revising this chapter; this will help you brush up on your concepts.

For this chapter, you should make a small formula notebook/flashcards and revise them weekly to keep them fresh in your mind.

The Sequence and Series Class 11 Notes prepared by Vedantu is helpful for students to score good marks in their board exams. These solutions are prepared based on important questions from the NCERT curriculum by the top faculty of Vedantu. The practice problems provided in CBSE Sequence and Series Notes will help students to revise the concepts and ace their exams. The solutions and concepts are prepared by experts to provide top-notch learning content to students. Experts have done a lot of research on the preparation of solutions to provide a unique and fun learning experience to students.

FAQs on Sequences and Series Class 11 Notes CBSE Maths Chapter 9 (Free PDF Download)

1. What are the 4 Types of Sequences?

The 4 types of sequences are:

Arithmetic Sequences

Geometric Sequences

Harmonic Sequences

Fibonacci Numbers

2. What is Sequence and Series?

A "sequence" is nothing but an ordered list of numbers. The numbers that are present in the ordered list are called as  "elements" or "terms" of the sequence. When you add up all the terms in a sequence, you get a "series"; the addition, as well as the resulting value, is called the "sum" or "summation." For example, the sequence "1, 2, 3, 4" contains the terms "1", "2", "3", and "4"; the corresponding series is the sum "1 + 2 + 3 + 4", and the series' value is 10.

3. How do I get the Free PDF of CBSE Class 11 Maths Notes Chapter 9 Sequences and Series?

Students can download the free PDF of CBSE Class 11 Maths Notes Chapter 9 Sequences and Series on Vedantu’s website which provides free PDF on different topics of Mathematics. The solutions and concepts are prepared by experts to provide top-notch learning content to students and is helpful for students to score good marks in their board exams.

4. Is Chapter 9 of Class 11 Maths tough?

Mathematics isn't as difficult as it seems. One can be a star in the CBSE Maths exam with excellent grades. You must solve all the questions from Chapter 9 of the Class 11 Maths NCERT textbook. You should also refer to Vedantu’s Revision Notes for Chapter 9 of Class 11 Maths. Mathematics can’t be mugged up as it depends upon your accuracy and precision. You should also solve previous years or sample papers available on Vedantu .

5. Which concepts are discussed in this chapter?

The relevance of sequences,' which play a key part in a range of human activities, is explained in chapter 9 Sequences and Series. In our daily lives, we encounter many examples of sequences, such as the human population, money placed in banks, the value of any product over some time, and so on. When a collection is arranged in such a way that its members are labeled as first, second, third, and so on, it is said to be listed in the form of a sequence.'

6. Write the first five terms of the sequences and obtain the corresponding series:

a1 = -1, an = an-1/n, n ≥ 2

Given, an = an-1/n and a1 = -1

a2 = a1/2 = -½

a3 = a2/3 = -⅙

a4 = a3/4 = -1/24

a5 = a4/5 = -1/120

Thus, the first 5 terms we obtained are -1, -½, -⅙, -1/24, -1/120.

7. Are the Revision Notes for Chapter 9 of Class 11 Maths important for the students?

The Revision Notes for Chapter 9 of Class 11 Maths are important for students as Class 11 is the nurturing and base for future competitive examinations. Good grades will result in direct admission to prestigious institutions. Most examinations are based on NCERT and going through Vedantu’s Chapter 9 of Class 11 Maths Revision Notes is important for a better score in the examination. You can easily find the Revision Notes online on Vedantu. These notes are accurate and reliable. These are also great when you want to do quick revisions before the exam.

8. From where I can download Revision Notes of Chapter 9 “Sequence and Series” of Class 11 Maths?

You can download Revision Notes of Chapter 9 “Sequence and Series” of Class 11 Maths from Vedantu’s official website (vedantu.com). On this site, there are notes for all the chapters of Class 11 Maths. Visit their website to download the notes that are available free of cost, as these will help you in revising the concepts in Chapter 9 which you’ve studied. These notes will save your time before the exams as you will have all the important points handy.

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NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series PDF Download

NCERT Solutions are essential for all students who want to score well in their exams. Our subject matter experts have developed these solutions to help the students score good marks in their examinations. These solutions are well-explained and include each and every topic. Another essential benefit of NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series is that it is created as per the latest pattern of CBSE (Central Board of Secondary Education). These solutions also ensure that a student gets a clear understanding of each and every concept. 

In the Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions, each and every type of question is included. From easy to hard, the students will get an idea of all the types of questions which can appear in the exam. Also, apart from this, the students will also get to know about the most common questions which have a high chance of appearing in the exam. 

It can be assumed that a student will score good marks in their exams if they will study the Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution on a regular basis. Our subject matter experts at Selfstudys have invested a good amount of time in creating them so that the students do not find any difficulty studying from these solutions. As the experts have been in the education line for quite some time, they are aware of the learning potential of every student, which is why they have created these solutions in a simple way so that no student faces any difficulty studying from them. Students can also use these solutions at the time of revision. 

The NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series can be easily downloaded from the website of Selfstudys i.e. Selfstudys.com. 

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series PDF 

The NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series PDF can be easily downloaded from the website of Selfstudys. As they are in PDF format, it becomes easier for the students to access them. They are also mobile-friendly. 

How to Download the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series? 

Downloading the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series is if you know the right steps. The steps to download them are as follows: 

• Visit the official website of Selfstudys i.e. Selfstudys.com.

• Once the website is completely loaded, you need to click on the three lines which you will find on the upper left side. Following that, you have to click on the ‘NCERT Books and Solutions’ column. 

• After clicking on the option of ‘NCERT Books and Solutions’, a new list will open where you need to select the option of ‘NCERT Solutions’. 

• Following that, you will be redirected to a page where you need to select the class and the subject for which you want to download the NCERT Solutions. 
• In this case, you need to select class 10 and the subject Maths to download the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series. 

What are the Features of the Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution? 

There are many features of the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series which can be very beneficial for all the students. The most important features include: 

• Well-Elaborated: The NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series is well-elaborated to make it easier for the students to understand all the concepts with in-depth knowledge. 
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• Well-organized Format: One thing which makes our Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution different from other solutions is that they are developed in a well-structured format. These solutions are organized systematically, following the order of the chapters and topics in the textbooks. 
• Accuracy and reliability: Our subject matter experts have invested a good amount of time in developing these NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series. They are accurate and reliable for the preparation of exams. These notes can also be helpful for the students who do self-study. 

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The NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series has several benefits to offer to the students. Some of the most important of them are 

• Clear Understanding of the Topic: The Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions provides in-depth knowledge of each and every topic which is present in the syllabus. Each chapter is explained in detail in a step-by-step process. To make learning easier and more interesting for all the students, diagrams and examples are also given. 
• Authentic Solutions: The Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution are completely authentic and can be trusted 100%. The highly qualified subject matter experts at Selfstudys have created these solutions in an authentic way which makes it simple for all the students to learn all the concepts. 
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• Conceptual Clarity: The NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series focuses on building a strong foundation for all the students. Students can get a deep understanding of the fundamental principles of maths formulas, which serve as a basis for further mathematical topics. 
• Boosts Confidence: If a student regularly goes through the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series, there is no doubt that the students will gain confidence in their maths power. This confidence can motivate them to explore more difficult topics and make a strong base for all of them. 

What are the tips to study from the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series? 

There are various tips that can help the students to study effectively from the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series: 

• Go through the chapter thoroughly: It is very important for all the students to first go through the chapter as it will help them to understand the basics. If their basics will get strong, they will automatically be able to grasp the other concepts in a better way. 
• Practice Problem solving step-by-step: It is advisable for all the students to solve the problems step-by-step which are present in the Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions without looking at the answers. This will allow them to develop problem-solving skills and test their understanding of the concepts. 
• Create a study schedule: It is important for students to create a study schedule and follow it consistently to stay focused. Devote at least 45 minutes to grasp the concepts thoroughly. Give enough time to each topic covered in NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series and review them daily. Sticking to the schedule is recommended to ensure timely completion of all the topics. 
• Do not get distracted: All the students should stay focused to concentrate better and should not get distracted. They should stay away from their mobile phones and other electronic gadgets, for example, TV, etc. to avoid all the distractions while studying Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution. 
• Practice Time Management: While doing exam preparation from the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series, it is advisable for all the students to practice time management by setting a time duration for solving problems. This will help you develop efficiency and improve your speed in solving problems accurately. 

How to Master the Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution ?

There are various ways to master NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series. Some of the most important of them are: 

• Study the NCERT Solutions: The students can start by studying the Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions. It is advisable for them to read the solutions carefully and understand them stepwise manner used to solve each problem. They should use them regularly as it will help them to understand the concepts in a clear manner. 
• Do regular practice of these solutions: It is advisable for all the students to regularly practice these NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series as it helps the students to do effective preparation for their exams. The conceptual knowledge of the students is also boosted by regularly practicing these solutions. 
• Do group discussions with your peers: It is advisable for all the students to do group discussions with their peers as discussing the concepts, problem-solving techniques, and solutions with others can help you understand them in a better way. Teaching and explaining concepts to others can deepen your understanding of all the concepts. 
• Revise regularly: Regularly revise the concepts and solutions from NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series. Review your notes, summaries, and solved problems. Revisit the solutions and attempt to solve them again without referring to the provided solutions. This revision will help reinforce your understanding and improve retention. 
• Take note of your mistakes and work on them: It is advisable for all the students to take note of their mistakes while solving the Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution and work on them. Find your weaknesses and identify the specific concepts or steps that caused confusion. This will help you improve your problem-solving skills. 

How to Maximise Your Learning Potential While Solving the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series? 

To maximize your learning potential while solving the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series, here are the tips which can help you: 

• Do a thorough reading of the chapter: All the students should read Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution thoroughly to maximize their learning potential as it will help them to understand the basic concepts which will make it easier for them to grasp all the concepts. 
• Practice regularly: One of the most important tips which every student should follow in order to maximize their learning potential is to practice these NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series regularly. It includes practicing the questions regularly in the Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions. This will help them in understanding the concepts with a deep understanding which can help them to score high marks in their exams. 
• Revise the concepts on a regular basis: It is very important for all the students to Revise all the concepts from the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series as it helps to recall all the information which they have studied during the preparation for the exam. You can also use different methods to revise the concepts on a regular basis, for example, making flashcards, notes, etc. 

What Are the Methods To Study From NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series? 

Studying for the exam requires focus and concentration. Below we will discuss various methods of how students can study for an Exam with the help of NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series: 

• Making a strong grip of basics: It is very important for all students to create a strong base for basics and learn them for NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series as it helps a student to get a clear understanding of the concepts. This makes it easier for the students to learn them fast. 
• Study the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series on a regular basis: It is advisable for all the students to study Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions on a regular basis as it will help them to do effective preparation for their exams and also increases the chances of the students to score good marks in their exams. 
• Always read the summary of your textbook: It is advisable for all students to always read the summary which is included in the Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution. The students should read them on a regular basis as it will help them to get a thorough knowledge of all the topics and will also help them to prepare well for the exams. This will increase their chances of scoring good marks in the exams. 
• Identify your strong and weak areas: It is very important for all the students to identify their strong and weak areas as it will help them to understand where they need to put more effort in order to do well in their exams. 
• Do timely revision: All the students should do timely revision from the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series if they want to do extremely well in their exams as it will ensure that you have created a strong base of all the topics in their minds. 

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Class 11 CBSE Applied Maths seq & Series ML AGGARWAL Exercise 6.1

Please select, q1. give an example of a seq which is not a progression.

Q2. If a seq is given by a 1 =2,a 2 =3+a 1 and a n =2a n-1 for n>2.Then,write the corresponding series upto 4terms

Q3. Write first 5 terms of the following seqs whose nth terms are given by (i) a n = 2n + 5 (ii) a n =n(n-1) (iii) a n =2 n (iv) a n =n 2 +1/2n-3

Q4. Find the indicated term(s) in each of the following seqs whose nth terms are : (i) a n = 4n - 3; a 17 , a 24 (i1) a n =(-1) n-1 n 3 ; a 9

Q5. Find the first five terms of each of the following seqs and obtain the corresponding series : (i) a 1 = 1, a n = a n-1 + 2, n ≥ 2 (1) a=3, a n =3a n-1 +2, for all n > 1

Q6. If the sum of n terms of a seq is given by S n = 2n 2 + 3n for all n e N, find the first 4 terms. Also find its 20th term.

Q7. First term of a seq is 1 and the (n + 1)th term is obtained by adding (n +1) to the nth term for all natural numbers n. Find the sixth term of the seq

• Math Article
• Sequences And Series Class 11

Sequences and Series Class 11

In  Sequences and Series class 11 chapter 9 deals with the study of sequences which follow a specific pattern called progression. In this chapter, the concepts such as  arithmetic progression (A.P),  geometric mean, arithmetic mean, the relationship between A.M. and G.M., special series in forms of the sum to n terms of consecutive natural numbers, sum to n terms of squares and cubes of natural numbers will also be studied.

Sequence and Series Class 11 Concepts

The topics and subtopics covered in chapter 9 – Sequences and Series class 11 concepts are:

• Introduction
• Arithmetic mean
• The general term of a G.P.
• Sum to n terms of a G.P.
• Geometric Mean (G.M.)
• Relationship Between A.M. and G.M.
• Sum to n Terms of Special Series

Sequences and Series Class 11 Notes

The different numbers occurring in any particular sequence are known as terms. The terms of a sequence are denoted by

a 1 , a 2 , a 3 ,….,a n

If a sequence has a finite number of terms then it is known as a finite sequence. A sequence is termed as infinite if it is not having a definite number of terms. The nth term of an AP is given by

a + (n-1) d.

Between any two numbers ‘a’ and ‘b’, n numbers can be inserted such that the resulting sequence is an Arithmetic Progression. A 1 ,  A 2 , A 3 ,……, A n  be n numbers between a and b such that a, A 1  ,  A 2 , A 3 ,……, A n , b is in A.P.

Here, a is the 1st term and b is (n+2)th  term. Therefore,

b = a + d[(n + 2) – 1] = a + d (n + 1).

Hence, common difference (d) = ( b-a)/(n+1)

Now, A 1 = a+d= a+((b-a)/(n+1))

A 2 = a+2d = a + ((2(b-a)/(n+1))

A n = a+nd= a + ((n(b-a)/(n+1))}

The nth term of a geometric progression is given by a n  = ar n-1

Sum of nth term:

where n = number of terms, a = first term and d = common difference

Video Lessons on Sequence and Series

A.p. & it’s properties.

H.P. And It’s Properties

Solved Examples on Sequence and Series

Q.1: If a n  = 2 n , then find the first five terms of the series.

Solution: Given: a n  = 2 n

On substituting n = 1, 2, 3, 4, 5, we get

a 1 = 2 1 = 2

a 2 = 2 2 = 4

a 3 = 2 3 = 8

a 4 = 2 4 = 16

a 5 = 2 5 = 32

Therefore, the required terms are 2, 4, 8, 16, and 32.

Q.2: Find the sum of odd integers from 1 to 2001.

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

It clearly forms a sequence in A.P.

Where, the first term, a = 1

Common difference, d = 2

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n – 2 = 2000

2n = 2000 + 2 = 2002

S n = 1001 x 1001

S n = 1002001

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Class 11 Maths

Sequence, Progression and Series Maths Notes

The word sequence is related to latin word ‘sequor’ which means follow. Thus simply sequence means to follow e.g. letters of English alphabet a, b, c, d ……… in a sequence.

Similarly if elements of any group are listed such that its elements are recognized as first, second or third etc. this arrangement is called sequence.

Hence, Set of numbers whose elements are arranged in a definite order under any rule, is called sequence, e.g. :

1. Numbers 1, 2, 3, 4, 5 ……….. form a sequence, since here difference of consecutive terms is 1 (one):

2. Similarly numbers 2, 4,6, 8 ……….. form a sequence. since here difference between consecutive terms is 2.

(iii) Numbers 4, 8, 12, 16 form a sequence, since here difference between consecutive terms is 4.

(iv) 6, 3, \(\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{3}{16}\), ………… also form a sequence, since here ratio between two consecutive terms is 2 : 1.

(v) \(\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}\) ………… is also a sequence, since difference between their reciprocals is a constant number.

(vi) \(\frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}\), ………… is als0 a sequence, since we see that sum or difference or ratio of two consecutive terms is according to some rule.

The sequence, which has finite number of terms is called finite sequence. e.g.,

Sequence 5, 10,15,20, 25 has 5 terms i.e., finite number. Thus this sequence is finite sequence. The sequence, which has not finite number of terms is called infinite terms. e.g., 2, 4. 6, 8, 10 ………… has infinite number of terms thus this sequence is called infinite sequence.

Note: From the above examples, it is clear that a sequence can be expressed as the following way :

1. Write some starting terms of sequence such that a rule appear to write the upcoming terms e.g., 1, 8. 27, ……… is a sequence whose general term is T n = n 3

2. By writing some starting terms of sequence their specification can be expressed e.g., in the sequence 2, 3, 5, 7, 11, ……… terms are prime numbers which is their speciality.

3. Write two terms of sequence and then other can he expressed as the form of earlier terms e.g., to express sequence 1, 4, 5, 9, 14….. We can write a 1 = 1, a 2 = 4 a n + 2 = a n + a n + 1 (n = 1, 2, 3 …….)

Sequence in Function Form

Here, a function f exist between set N and A which has unique image in A for each element N

f(1) = a 1 f(2) = a 2 , f(3) = a 3 , etc. Here domain of f is N (Set of natural numbers) and range of f is A (Set of terms of sequence) This sequence in function form is written as f(a). Thus, f(a) = a 1 , a 2 , a 3 , ………. , a n Now If we consider numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ………. then we see that they do not represent any order but we reconstruct it. As :

1 st term = 2 nd term = 1 3 rd term = 2 = 1 st term + 2 nd term = 1 + 1 = 2 3 rd term 4 th term = 2 nd term + 3 rd term 3 = 1 + 2 5 th term = 3 rd term + 4 th term 5 = 2 + 3

If 1 t , 2 nd , 3 rd terms are represented a 1 , a 2 , a 3 , a 4 , ……….. etc Then, a 1 = a 2 = 1 a 3 = a 1 + a 2 = 1 + 1 = 2 a 4 = a 2 + a 3 = 1 + 2 = 3 a 5 = a 3 + a4 = 2 + 3 = 5 …………………………….. …………………………….. or a n = a n – 2 + a n – 1 , whereas n > 2 Sequence 1, 1, 2, 3, 5, 8 ……………… is called Fibonacci sequence.

If a 1 , a 2 , a 3 , a 4 , …………, a n …. is a sequence, then expression a 1 ± a 2 ± a 3 ± a 4 ± ……. ± a n ± ……….. is called series. Therefore there is a series, corresponding to each sequence, in which two terms are joined with positive or negative sign. Each series has a corresponding sequence.

Progression

Progression : A sequence is called progression if numeric value of its terms increase or decrease under some specific rule. i.e., n th term of a sequence can be find by a formula then it is called progression.

Difference between Sequence and Series: By sequence we mean an arrangement of numbers in definite order according to some rule. It is not always possible to write n th term in sequence but in series we always write n th term.

For example: In the sequence of 2, 3, 5, 7, 11, 13 ………… we find that there is no formula for the n th term.

Arithmetic Progression

A sequence a 1 , a 2 , a 3 , a 4 , …….., a n – 1 , a n is called Arithmetic Progression. If d = a n – a n – 1 n ∈ N where a 1 is first term of progression and d is common difference of Arithmetic Progression.

If we consider natural numbers 1, 2, 3, 4, 5, ……….. then we see that 2 – 1 = 1, 3 – 2 = 1, 4 – 3 = 1, 5 – 4 = 1 etc. i.e., difference of any two consecutive natural numbers is 1. Thus, sequence in which difference between two consecutive terms remains same, is called Arithmetic Progression.

If a is first term of any sequence, d their common difference then 1 st , 2 nd , 3 rd , 4 th , 5 th terms of Arithmetic progression will be a + d, a + 2d, a + 3d, a + 4d, ………..

a, a + d, a + 2d, a + 2d is called general Arithmetic Progression. Here by putting different values of a and d we can get different Arithmetic Progressions. In brief, Arithmetic Progression is called A.P..

Characteristics of Arithmetic Progression

• If same constant is added to each term of any A.P then sequence so obtained will also be A. P.
• If same constant is subtracted from each term of any A.P, then sequence so obtained will also be A.P.
• If same constant is multiplied to each term of any A.P, then sequence so obtained will also be A.P.
• If each term of any A.P. is divided by any non-zero constant then sequence so obtained will also be A.P.

General Term of A.P.

In any A.P, n th term from start is called its General Term and is denoted by T n .

If a is first term of any r..?. and d is their common difference, then general term or (n th term) T n = a + (n – 1) d

Last Term of A.P.: If in a A.P, first term is a, common difference is d then last term l = a + (n – 1) d, where n is number of terms in A.P.

In A.P, nth term from last: If in a A.P, a is first term, common difference d and last term is l, then n th term from last = l – (n – 1) d

Sum of first ‘n’ Terms of an A.P.

Sum of first n terms of A.P. ¡s denoted by S n . Let first term of given A.P. is a, common difference is d and n th term is l. a, a + d, a + 2d ……. 1 – 2d, 1 – d + l will be terms of progression. Thus S n = a + (a + d) + (a + 2d) + …….. + (l + 2d) + (l – d) + l ……… (i) S n = l + (l – d) + (l – 2d) + …….. + (a + 2d) + (a + d) + a ……… (ii) Adding corresponding term of (i) and (ii) 2S n = (a + l) – (a + l) + (a + l) + ……….. + (a + l) + (a + l) ………. (n term) = n(a + l) ∴ S n = \(\frac{1}{2}\) n(a + l) S n = \(\frac{1}{2}\) [a + a + (n – 1)d] [∵ l = T n = a + (n – 1)d] or S n = \(\frac{1}{2}\) [2a + (n – 1)d]

Properties of A.P.

1. If in each term of any A.P a definite number is add or subtract, then obtained progression will be A.P with same common difference.

2. If each term fo any A.P is multiply or divide by a certain non zero number, then obtained progression will be A.P.

3. In any finite A.P., sum of same distant terms from start and end remains constant and is equal to sum of first and last term.

4. Each term of any A.P (except first and last term) is half of the sum of two terms lie on same distance.

5. If in any A.P, number of terms is odd, then sum of this series is equal to the product of middle term and number of terms.

6. It x 1 , x 2 , ….x n and y 1 , y 2 , …. y n , are two arithmetic progression of n terms, then (x 1 ± y 1 ), (x 2 ± y 2 ), (x 3 ± y 3 ) …………. (x n ± y n ) will be A.P

Geometric Progression (G.P.)

Thus this ratio remains constant. This ratio is denoted by r. By multiplying constant r with any term we get another term.

Definition of a Geometric Progression

Sequence of numbers, in which each term is obtained by multiplying previous number by a definite number (constant number), is called Geometric Progression.

Geometric Progression is called GP., (in brief).

Let first term of any sequence a 1 = a and its common ratio is r, then 1 st term × common ratio = r × a = ar = a 2 , 2 nd term of sequence Similarly, 2nd term × common ratio r = ar × r = ar 2 = a 3 , 3rd term of sequence and 3rd term × Common ratio = ar 2 × r = ar 3 = a 4 , 4 th term of sequence Thus, sequence (a 1 , a 2 , a 3 , a 4 , ………..) = a, ar, ar 2 , ar 3 , …… represents a G,P. i.e., First term a and common ratio is r of GP. a, ar, ar 2 , ar 3 , ar 4 , …….. ar k – 1 …

General Term of G.P

Geometric Mean, G.M.

To Find n Geometric Mean Between Two Numbers

Relationship between A.M. and G.M.

it is clear that A – G > 0 or A > G Thus, A.M. between two numbers > GM.

Important Properties & AM. and G.M. between Two Quantities

Property 2. If A and G are A.M. and GM. respectively between two positive quantities a and b then quadratic equation with roots a and b is i.e., x 2 – 2Ax + G 2 = 0. Proof: Here A = \(\frac{a+b}{2}\) and G = √4ab ….(i) Quadratic equation with roots a and b is x 2 – (a + b)x + ab = 0 or x 2 – 2Ax – G 2 = 0 [Using (i)]

Sum of an infinite G.P.

In this type of sequence number of terms is infinite as : 1, \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}\), ……. is an infinite GP., in which first term 1 and common ratio is 1/3. If |r| < 1 then sum of infinite series S ∞ = \(\frac{a}{1-r}\)

Note : If |r| > 1. then the sum of their terms is infinite. If |r| < 1, then as terms increases, value of term decreases. It is possible to find any limitation. So that the sum of G.P. is not too much. If no. of terms are too much, then tire sum of GP. comes near to limitation. In this case, this limit is called sum of infinite GP.

Recurring Decimals

The decimals, in which after decimal any digit or a definite group of digits repeated is called recurring decimals. To represent recurring decimal sign (∙) is put on recuring digit or group of digits as : 6666… = 6 0.153535353… = 0.153 etc.

Recurring decimals are example of such G,P, whose sum of infinite terms is finite number. With the help of GP., value of recurring decimal can be found. For this purpose formula S ∞ = \(\frac{a}{1-r}\) is used.

Arithmetic Geometric Series

By multiplying corresponding terms of A.P. and GP., series so obtained is called Arithmetic-Geometric Series. For example : 1 + 3x + 5x 2 + 7x 3 + ……. is a Arithmetic Geometric Series, in which 1, 3, 5, 7 …… are in A.P. and 1, x, x 2 …… are in GP. n th term of this A.P. is (2n – 1) and n th term of G.P. is x n – 1 . Thus n th term of Arithmetic Geometric Series will be (2n – 1)x n – 1 . General form of Arithmetic Geometric Series is a, (a + d) r, (a + 2d) r 2 , …….. and its general term is T n = {a + (n – 1 )d}r n – 1 .

To find sum of first n terms of Arithmetic Geometric Series:

Let sum of first n terms of any A.G.P. is S n . Then S n = a + (a + d)r + (a + 2d) r 2 + (a + 3d) r 3 +…. + [a + (n – 1 )d]r n – 1 ……(i)

Note: Sum should be find by above given method not by using the above results (iii) and (iv).

Sum of Arithmetic Geometric Series by Difference Method

If in a series difference between consecutive term is in G.P., then to find sum of this series, under the terms of given series. We write terms of this series by extend one by one and then substract, obtained terms will be in GP. So we can find n th term of series and by putting n = 1,2…. We get each term and then we can find sum of series by addition.

Sigma Notation

Sum to n Terms of Series of Natural Numbers, their Squares and Cubes

Sum of Series by Difference Method

If difference between consecutive terms in any series is in A.P., then to find sum of this series, under the terms of given series we write terms f this series by extend one by one then subtract. Obtained terms will be in A.P. So we can find nth term of series and we can find sum of series by using formula of Σn. Σn 2 and Σn 3 .

Harmonic Progression

If reciprocals of each term of a series, in same order are in A.P., then this series is called Harmonic Progression. Consider the following Progression

• \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)
• \(\frac{1}{20}, \frac{1}{17}, \frac{1}{14}, \frac{1}{11}, \ldots\)

Above progressions are harmonic progressions because their reciprocals 3, 5, 7, 9 …. and 20, 17, 14, 11 …… are in A.P.

General Term of H.P.

Note : 1. To solve questions of H.P. take reciprocals serially of each term and make A.P. corresponding to this and then solve A.P. by using related formulas. 2. There is no general formula to Find sum of n terms of H.P.

Harmonic Mean

If three or more than three numbers are in H.P., then all the numbers between first and last number are called Harmonic Mean between them.

For example, If a, H 1 , H 2 , H 3 , ……… H n b are in H.P., then H 1 , H 2 , H 3 , ……… H n , are called n H.M. between a and b. Harmonic Mean, in brief expressed as H.M.

To Find One H.M. between Two given Numbers

To find n Harmonic Mean between Two Given Numbers

Note: From above it is clear that to find H.M. first we will find A.M. of corresponding A.P. Reciprocals of obtained A.M. will be required Harmonic means. n th term of H.P. H.P. = \(\frac{a b(n+1)}{n a+b}\) and In general form, r th term H r = \(\frac{a b(n+1)}{(n-r+1) b+r a}\)

Relation between AM, G.M. and H.M

Theorem (i) GM, of two real positive numbers is geometric mean of their A. M. (A) and H.M. (H) i.e., AH = G 2

Condition for Three Quantities to be in A.P., G.P. and H.P.

Maths Notes

Sequences and Series Class 11 Notes Maths Chapter 9

Class 11 Maths Notes students can refer to Sequences and Series Class 11 Notes Maths Chapter 9 https://www.cbselabs.com/sequences-and-series-class-11-notes/ Pdf here. They can also access the CBSE Class 11 Sequences and Series Chapter 9 Notes while gearing up for their Board exams.

CBSE Class 11 Maths Notes Chapter 9 Sequences and Series

Sequence And Series Class 11 Notes Chapter 9

Sequence A succession of numbers arranged in a definite order according to a given certain rule is called sequence. A sequence is either finite or infinite depending upon the number of terms in a sequence.

Infinite Series Calculator is a free online device that gives the summation estimation of the given capacity for as far as possible.

Series If a 1 , a 2 , a 3 ,…… a n is a sequence, then the expression a 1 + a 2 + a 3 + a 4 + … + a n is called series.

Progression A sequence whose terms follow certain patterns are more often called progression.

Arithmetic Progression (AP) A sequence in which the difference of two consecutive terms is constant, is called Arithmetic progression (AP).

Sequences And Series Class 11 Notes Chapter 9

Properties of Arithmetic Progression (AP) If a sequence is an A.P. then its nth term is a linear expression in n i.e. its nth term is given by An + B, where A and S are constant and A is common difference.

nth term of an AP : If a is the first term, d is common difference and l is the last term of an AP then

• nth term is given by a n = a + (n – 1)d.
• nth term of an AP from the last term is a’ n =a n – (n – 1)d.
• a n + a’ n = constant
• Common difference of an AP i.e. d = a n – a n-1 ,∀ n > 1.

If a constant is added or subtracted from each term of an AR then the resulting sequence is an AP with same common difference.

If each term of an AP is multiplied or divided by a non-zero constant, then the resulting sequence is also an AP.

If a, b and c are three consecutive terms of an A.P then 2b = a + c.

Any three terms of an AP can be taken as (a – d), a, (a + d) and any four terms of an AP can be taken as (a – 3d), (a – d), (a + d), (a + 3d)

Find the sequence calculator – find sequence types, indices, sums and progressions step-by-step.

Class 11 Sequence And Series Notes Chapter 9

Sum of n Terms of an AP Sum of n terms of an AP is given by S n = \(\frac { n }{ 2 }\) [2a + (n – 1)d] = \(\frac { n }{ 2 }\) (a 1 + a n )

A sequence is an AP If the sum of n terms is of the form An 2 + Bn, where A and B are constant and A = half of common difference i.e. 2A = d.

a n =S n – S n-1

Class 11 Maths Chapter 9 Notes

Arithmetic Mean If a, A and b are in A.P then A = \(\frac { a+b }{ 2 }\) is called the arithmetic mean of a and b.

The sum of a linear number sequence calculator adds up numbers in a linear sequence.

Geometric Progression (GP) A sequence in which the ratio of two consecutive terms is constant is called geometric progression. The constant ratio is called common ratio(r). i.e. r = \(\frac { { a }_{ n }+1 }{ { a }_{ n } }\), ∀ n > 1

Sequence And Series Notes Chapter 9 Class 11

Properties of Geometric Progression If a is the first term and r is the common ratio, then the general term or nth term of GP is a n =ar n-1

nth term of a GP from the end is a’ n = \(\frac { 1 }{ { r }^{ n-1 } }\), l = last term

If all the terms of GP be multiplied or divided by same non-zero constant, then the resulting sequence is a GP with the same common ratio.

The reciprocal terms of a given GP form a GP.

If each term of a GP be raised to some power, the resulting sequence also forms a GP

If a, b and c are three consecutive terms of a GP then b 2 = ac.

Any three terms can be taken in GP as \(\frac { a }{ r }\), a and ar and any four terms can be taken in GP as \(\frac { a }{ { r }^{ 3 } }\), \(\frac { a }{ r }\), ar and ar 3 .

Sequence And Series Class 11 Notes Pdf Chapter 9

This Limit of Sequence Calculator handy tool is easy to use and provides the steps for easy understanding of the topic.

Geometric Mean (GM) If a, G and b are in GR then G is called the geometric mean of a and b and is given by G = √(ab).

If a,G 1 , G 2 , G 3 ,….. G n , b are in GP then G 1 , G 2 , G 3 ,……G n are in GM’s between a and b, then common ratio r = \(\left( \frac { b }{ a } \right) ^{ \frac { 1 }{ n+1 } }\)

If a 1 , a 2 , a 3 ,…, a n are n numbers are non-zero and non-negative, then their GM is given by GM = (a 1 . a 2 . a 3 …a n ) 1/n

Product of n GM is G 1 × G 2 × G 3 ×… × G n =G n = \(\left( ab \right) ^{ \frac { n }{ 2 } }\)

Notes Of Sequence And Series Class 11 Chapter 9

Important Results on the Sum of Special Sequences Sum of first n natural numbers is Σn = 1 + 2 + 3 +… + n = \(\frac { n(n+1) }{ 2 }\)

Sum of squares of first n natural numbers is Σn 2 = 1 2 + 2 2 + 3 2 + … + n 2 = \(\frac { n(n+1)(2n+1) }{ 6 }\)

Sum of cubes of first n natural numbers is Σn 3 = 1 3 + 2 3 + 3 3 + .. + n 3 = \(\left( \frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } \right) ^{ 2 }\)

Sequence And Series Notes Class 11 Chapter 9

Notes Of Maths Class 11 CBSE Chapterwise

• Chapter 1  Sets Class 11 Notes
• Chapter 2 Relations and Functions Class 11 Notes
• Chapter 3 Trigonometric Functions Class 11 Notes
• Chapter 4 Principle of Mathematical Induction Class 11 Notes
• Chapter 5 Complex Numbers and Quadratic Equations Class 11 Notes
• Chapter 6 Linear Inequalities Class 11 Notes
• Chapter 7 Permutations and Combinations Class 11 Notes
• Chapter 8 Binomial Theorem Class 11 Notes
• Chapter 9 Sequences and Series Class 11 Notes
• Chapter 10 Straight Lines Class 11 Notes
• Chapter 11 Conic Sections Class 11 Notes
• Chapter 12 Introduction to Three Dimensional Geometry Class 11 Notes
• Chapter 13 Limits and Derivatives Class 11 Notes
• Chapter 14 Mathematical Reasoning Class 11 Notes
• Chapter 15 Statistics Class 11 Notes
• Chapter 16 Probability Class 11 Notes

NCERT Solutions

• School Guide
• Class 11 Syllabus
• Class 11 Revision Notes
• Maths Notes Class 11
• Physics Notes Class 11
• Chemistry Notes Class 11
• Biology Notes Class 11
• NCERT Solutions Class 11 Maths
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Geometric Series

• Absolute Value of a Complex Number
• Statements - Mathematical Reasoning
• Estimating Limits from Graphs
• Estimating Limits from Tables
• Arithmetic Progression and Geometric Progression
• Average and Instantaneous Rate of Change
• Applications of Power Rule
• Formal Definition of Limits
• Strategy in Finding Limits
• Algebra of Derivative of Functions
• Focus and Directrix of a Parabola
• Derivatives of Trigonometric Functions
• Identifying Conic Sections from their Equation
• Limits by Direct Substitution
• Cartesian Product of Sets
• Standard Form of a Straight Line
• Properties of Limits
• Sequences and Series

Sequence and Series Word Problems | Class 11 Maths

Sequences and series have several important applications in several spheres of human activities. When sequences follow some specific patterns, they are usually called progressions. Arithmetic and Geometric progressions are some examples of commonly occurring progressions. Let’s see some problems on these progressions to understand them better. 

Growth Patterns

Such types of number sequence problems first describe how a sequence of numbers is generated. Some terms of the sequence are given, and we need to figure out the patterns in them and then the next terms of the sequence. 

Solving such sequences: 

• Look for a pattern between the given numbers.
• Decide whether to use +, -, × or ÷
• Use the pattern to solve the sequence.

Increasing Pattern

Increasing Patterns as the name suggests will always be increasing in nature. The next term and the term just before that will all be similarly related with the help of operations (×, -, +).

Question: 6,13,27,55, ….. In the given sequence, what is the value of the next term? 

On looking carefully into the pattern, one can see that 13 = 6 × 2 + 1  27 = 13 × 2 + 1  55 = 27 × 2 + 1  This shows that every term is twice the preceding term plus one. So, let the next term be “a”.  a = 55 × 2 + 1     = 110 + 1     = 111  Hence, the next term is 111. 

Decreasing Pattern

In decreasing Patterns, the next will be lesser than the previous term and two consecutive terms will follow a certain pattern.

Question: What is the next term in the series: 220,100,40, ….

Answer:  

The series follow the pattern: 100 = (220×0.5)-10                                                 40 = (100×0.5)-10 Therefore, the next term will be (40×0.5)-10 = 10

Arithmetic Progressions 

In Arithmetic Progressions, the consecutive terms will have the same difference and are denoted as ‘d’, the first term is called as ‘a’, and the number of terms is denoted as ‘n’. 

The formula for n th term is: T n = a+ (n-1)d

Question 1: 2, 5, 8, 11, …. Find the next term of the sequence. 

Answer: 

It can be noticed by carefully studying the terms of the sequence that the difference between each consecutive term remains the same. For example:  5 – 2 = 3  8 – 5 = 3 11 – 8 = 3  So, the next will be at a difference of three from the last term. Since the last term of the sequence is 11. The next terms will be 14. OR The formula for n th term in Arithmetic progression can also be used here, 5 th term is required here: a = 2 d = (3) T 5 = 2+(5-1)(3)  = 2+12 = 14

Question 2: 15, 12, 9, … __. Find the next term.

In this problem also, all the terms have a difference of three between them. The difference is that the sequence in decreasing in nature. Since the last term is 9, the next term will be 3 less than the last term. So, the last term will be 6. 

Fibonacci Sequence

Sometimes there are sequences for which pattern is not visible, the Fibonacci sequence is an example of such a sequence. It is a very commonly occurring sequence in the field of mathematics and computer science.

The number is arranged as, 1, 1, 2, 3, 5, 8 …. Here the pattern is not visible, this sequence proceeds in a manner that depends on its history. 

Let a n be the nth term for the sequence. In this sequence a n = a n-1 + a n-2 . 

For example, 

a 2 = a 1 + a 0 i.e 2 = 1 + 1,  a 3 = a 1 + a 2 i.e 3 = 2 + 1,  a 4 = a 3 + a 2 i.e 5 = 2 + 3 and,  a 5 = a 4 + a 3 i.e 8 = 5 + 3. 

Question: What is the next term in the Fibonacci series: 1,1,2,3,5,8,13,……

In Fibonacci Series, the next term is the sum of the last two terms. Therefore, the next term will be (8+13)= 21

Before starting out with the problems related to a geometric progression. Let’s take a quick recap of the formulas for the sum and nth term of a GP. 

The general form of a geometric progression is a, ar, ar 2 , ar 3 …… where a = first term, r = common ratio and a n be the nth term. 

• The nth term of the progression: a n = ar n-1 .

Finite GP problems

These kinds of problems include the geometric series where there are a finite number of terms. 

Question 1: The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour, and nth hour? 

The growth of the bacteria makes a GP, 30, 60, 120,….. and so on.  In this GP, a = 30, r = 2. Let the number of bacteria at nth hour be a n .  At n = 2, a 2 = ar 2-1                       = (30)(2) 2-1 = 30 × 2 = 60  At n = 4, a 4 = ar 4-1                       = (30)(2) 4-1 = 30 × 2 3 = 240 At nth step an = ar (n-1) = 30 × 2 n-1

Question 2: A person has 2 parents, 4 grandparents, 8 great-grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own. 

This is a problem of finite GP. The sequence can be thought like this,  2, 4, 8, 16, …..  So, the total number of ancestors in 10 generation of his family.  2, 4, 8, 16, 32, …..10 terms.  Here, a = 2, r = 2 and n = 10  S10 = a 

Infinite GP Problems

An infinite geometric series is the sum of an infinite geometric sequence. This series would have no last term

Question 1: A monkey is swinging from a tree. On the first swing, she passes through an arc of 24m. With each swing, she passes through an arc of 24m. With each swing, she passes through an arc half the length of the previous swing. What is the total distance traveled by the monkey when she completes her 100000th swing? 

Now this movement represents a GP with a = 24 and r = 1/2. Now, since the GP is decreasing and the question asks for the sum till 100000th term. To save the calculation , we can consider it an infinite GP and round of the answer we get.  Sum of an infinite GP =  Here, a = 24 and r = 1/2. Let the sum be S  So the monkey travels almost 24m in these many swings. 

Question 2: A ball was dropped from a 24-inch high table. The ball rebounds and always reaches three fourth of the distance fallen. What is the approximate distance the ball travels before finally coming at rest on the ground. 

It should be noticed that this problem actually involves two infinite geometric series. The first series involves ball falling and the other series involves ball rising after rebounding from the ground.  Falling: a 1 = 24 , r = 3/4  Rising: a 2 = 24(3/4) = 18 ,r = 3/4  Using the formula for infinite geometric series,  S =  Let S be the total distance travelled: S = S rising + S falling S rising = S falling =  Now, S = S rising + S falling = 72 + 96 = 168

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